How is Schwartz space different from Hilbert space? I know that Schwartz space can be considered a dense subset of the Hilbert space isomorphic to $\ell^2$. What I wish to understand is, how really different Schwartz space is from the Hilbert space.
Schwartz space has completeness, and is an inner product space, at least understood as "borrowing" inner product structure from the Hilbert space. This seems to satisfy all the definitions of the Hilbert space. Yet we know that these two spaces are not isomorphic. So what distinguish them?
Maybe the problem is that inner product borrowed from the Hilbert space is not natural to the natural topology of Schwartz space? It would be nice if someone clarifies on this. 
And I believe quantum mechanics uses inner product on Schwartz space, so I guess Schwartz space is understood at least there as an inner product space...
 A: Let $\mathscr{S}$ denote the Schwartz space. We equip this space with a topology by giving it the family of seminorms
$$
p_{\alpha, \beta}(f) := \sup_{x \in \mathbb{R}^n} |x^\beta \partial^\alpha f(x)|
$$
where $\alpha,\beta$ are multi indices. With respect to the induced topology, we know that $\mathscr{S}$ is completely metrizable, as you have noted. 
On the other hand, one can view $\mathscr{S}$ as a subspace of $L^2(\mathbb{R}^n)$. Hence, $\mathscr{S}$ inherits an inner-product (and a metric topology). However, this is not the same topology as above! Indeed, $\mathscr{S}$ is not even complete with respect to the metric inherited from $L^2(\mathbb{R}^n)$. This is because $\mathscr{S}$ is not closed as a subspace of $L^2(\mathbb{R}^n)$. To see this, simply note that $$\overline{\mathscr{S}} = L^2(\mathbb{R}^n) \neq \mathscr{S}$$ when interpreted as a subspace of $L^2(\mathbb{R}^n)$.
A: Since you mentioned $\ell^2$ and quantum mechanics, here is a clean way to see the difference between the Schwartz space $\mathscr{S}(\mathbb{R})$ and $L^2(\mathbb{R})$.
Take the basis of Hermite functions $(\psi_n)_{n\ge 0}$ which diagonalizes the Hamiltonian for the harmonic oscillator $H=\frac{1}{2}(P^2+X^2)$. This gives an isomorphism $T:L^2(\mathbb{R})\rightarrow \ell^2$, $f\rightarrow \langle \psi_n,f\rangle_{L^2}$.
The image of $\mathscr{S}(\mathbb{R})\subset L^2(\mathbb{R})$ is the space $\mathfrak{s}$ of sequences $x=(x_n)_{n\ge 0}$ with rapid decay. These are the sequences such that for all $k\ge 0$,
$$
||x||_k=\sup_{n\ge 0}\ (n+1)^k|x_n|
$$
is finite.
In fact, the map $T$ realizes a topological vector space isomorphism between $\mathscr{S}(\mathbb{R})$ and $\mathfrak{s}$ if the latter is equipped with the topology defined by the seminorms $||\cdot||_{k}$, $k\ge 0$.
